Newton-Raphson Method

If you've ever tried to find a root of a complicated function algebraically, you may have had some
difficulty. Using some basic concepts of calculus, we have ways of numerically evaluating roots of
complicated functions. Commonly, we use the Newton-Raphson method. This iterative process
follows a set guideline to approximate one root, considering the function, its derivative, and an
initial x-value.

You may remember from algebra that a root of a function is a zero of the function. This means that
at the "root" the function equals zero. We can find these roots of a simple function such as:
f(x) = x2-4 simply by setting the function to zero, and solving:

f(x) = x2-4 = 0
(x+2)(x-2) = 0
x = 2 or x = -2

The Newton-Raphson method uses an iterative process to approach one root of a function. The specific root
that the process locates depends on the initial, arbitrarily chosen x-value.

Here, xn is the current known x-value, f(xn) represents the value of the function
at xn, and f'(xn) is the derivative (slope) at xn. xn+1
represents the next x-value that you are trying to find. Essentially, f'(x), the derivative
represents f(x)/dx (dx = delta-x). Therefore, the term f(x)/f'(x) represents a value of dx.

The more iterations that are run, the closer
dx will be to zero (0). To see how this works, we will perform the Newton-Raphson method on the function that
we investigated earlier, f(x) = x2-4. Below are listed the values that we need to know
in order to complete the process.

Theoretically, we could execute an infinite number of iterations to find a perfect representation for
the root of our function. However, this is a numerical method that we use to lessen the burden
of finding the root, so we do not want to do this. Therefore we will assume that the process has
worked accurately when our delta-x becomes less than 0.1. This value of precision should be specific
to each situation. A much more, or a much less, precise value may be appropriate when using the Newton-Raphson
method in class. The table below shows the execution of the process.

n

xn

f(xn)

f'(xn)

xn+1

dx

0

x0 = 6

f(x0 = 32)

f'(x0 = 12)

x1 = 3.33

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1

x1 = 3.33

f(x1) = 7.09

f'(x1) = 6.66

x2 = 2.27

dx = 1.06

2

x2 = 2.27

f(x2) = 1.15

f'(x2) = 4.54

x3 = 2.01

dx = .26

3

x3 = 2.01

f(x3) = 0.04

f'(x3) = 4.02

x4 = 2.00

dx = 0.01

Thus, using an initial x-value of six (6) we find one root of the equation
f(x) = x2-4 is x=2. If we were to pick a different inital x-value, we may find the same
root, or we may find the other one, x=-2.

A graphical representation can also be very helpful. Below, you see the same function
f(x) = x2-4 (shown in blue). The process here is the same as above. In the first
iteration, the red line is tangent to the curve at x0. The slope of the tangent is the
derivative at the point of tangency, and for the first iteration is equal to 12.
Dividing the value of the function at the initial x (f(6)=32) by the slope of the tangent (12), we
find that the delta-x is equal to 2.67. Subtracting this from six (6) we find that the new x-value
is equal to 3.33. Another way of considering this is to find the root of this tangent line. The
new x-value (xn+1) will be equal to the root of the tangent to the function at the current x-value
(xn).

The Newton-Raphson method does not always work, however. It runs into problems in several places. First,
consider the above example. What would happen if we chose an initial x-value of x=0? We would have
a "division by zero" error, and would not be able to proceed. You may also consider operating
the process on the function f(x) = x1/3, using an inital x-value of x=1. Do the x-values converge? Does the delta-x
decrease toward zero (0)?

So, how does this relate to chemistry? Consider the van der Waals equation found in the Gas Laws
section of this text. Assuming that we have a set number of moles of a set gas, not under ideal
conditions, we can use the Newton-Raphson method to solve for one of the three variables (temperature, pressure,
or volume), based on the other two. To do this, we need to use the van der Waals equation, and the
derivative of this equation, both seen below.

As you can see, the Van der Waals equation is quite complex. It is not possible to solve
it algebraically, so a numerical method must be used. The Newton-Raphson Method is the easiest
and most dependable way to solve equations like this, even though the equation and its derivative
seem quite intimidating.

Depending on the conditions under which you are attempting to solve this equation, several of the
variables may be changing. So, it may be necessary to use partial derivatives. For the purposes
of this example, we are assuming that pressure, temperature, and volume are the only things changing,
and that these values are all functions of time. This avoids the use of a partial derivative; we simply
differentiate all variables with respect to time, as shown above. Some algebraic manipulation of the
equation and/or its derivative may be needed depending on the specific problem to be solved. It is
assumed that all of the variables but one are specified; that variable is used in the expression for
"xn+1" that Newton's method uses. Performing Newton's method on this equation successfully
would give a value of that variable which gives a solution when the other variables are held constant at
the values you specified.